In a series R, L, C circuit `X_(L)=10Omega, X_(c) = 4omega`and `R = 6Omega`. Find the power factor of the circuit

Options

A. `(1)/sqrt2`
B. `(sqrt3)/(2)`
C. `1//2`
D. nonc of theco

Answer

To find the power factor of the given series RLC circuit, we will follow these steps:

### Step 1: Identify the given values
We have the following values:
- Inductive reactance, \( X_L = 10 \, \Omega \)
- Capacitive reactance, \( X_C = 4 \, \Omega \)
- Resistance, \( R = 6 \, \Omega \)

### Step 2: Calculate the net reactance
The net reactance \( X \) in a series RLC circuit can be calculated using the formula:
\[ X = X_L - X_C \]
Substituting the given values:
\[ X = 10 \, \Omega - 4 \, \Omega = 6 \, \Omega \]

### Step 3: Calculate the impedance \( Z \)
The total impedance \( Z \) in the circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values of \( R \) and \( X \):
\[ Z = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \, \Omega \]

### Step 4: Calculate the power factor
The power factor \( \text{PF} \) is given by the formula:
\[ \text{PF} = \frac{R}{Z} \]
Substituting the values of \( R \) and \( Z \):
\[ \text{PF} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Conclusion
Thus, the power factor of the circuit is:
\[ \text{PF} = \frac{1}{\sqrt{2}} \]